2019 A/L Physics Past Paper · Question 24 (English Medium)

Answer: (5) Diode 20 Ω, resistor 20 Ω

What the question asks

What we see:

One way, $1$ V gives $50$ mA.

The other way, $1$ V gives $100$ mA.

We want the connection inside the box, the forward resistance of the diode $r_d$ and the resistor $R$.

Relevant theory

A diode conducts when it is forward biased, with a resistance $r_d$.

It blocks the current when it is reverse biased.

So the current changes when we turn the voltage round because of the diode.

Unlike a normal metal junction, p - n junction allows a current to flow when it is forward biased and does not allow a current to flow when it is reverse biased.

Required equations

$I = \frac{V}{R_{eq}}$

$\frac{1}{R_{eq}} = \frac{1}{R} + \frac{1}{r_d}$

Reading the options

The options are pairs $(r_d, R)$.

We find the connection first.

Then we look for the pair that fits both readings.

Reasoning and calculation

Two clues:

The current flows both ways. So the diode is not in series with the resistor. In series the current would be zero one way.

One way the current doubles. So an extra path opens in that way.

So the diode is in parallel with the resistor.

In the first way the diode is reverse biased. The current goes through $R$ only.

In the other way the diode conducts too. Both paths carry current.

$20$

$R = \frac{V}{I_1} = \frac{1}{0.05}$

$0.05$

$I_d = I_2 - I_1 = 0.1 - 0.05$

$r_d = \frac{V}{I_d} = \frac{1}{0.05}$

If $r_d = 0$ the current has no limit.

With $R = 10\ \Omega$ the first way gives 100 mA.

With $r_d = 10\ \Omega$ the turned round way gives 150 mA.

Only one pair fits both readings.

It is $r_d = 20\ \Omega$ and $R = 20\ \Omega$.

That is option (5).

Checks

$10$

$R_{eq} = \frac{20 \times 20}{20 + 20}$

$0.1$

$I_2 = \frac{1}{10}$

Conclusion

$r_d = 20\ \Omega,\ R = 20\ \Omega$

The resistor is $20\ \Omega$. It alone gives 50 mA when the diode blocks.

Turned round, the diode of $20\ \Omega$ conducts too, in parallel.

The current is 100 mA.

So the answer is option (5).